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I was reading something when the author said that "x converges to a dirac delta function". Was wondering if someone could explain what it means. I work in IB and am unsure what this means.

The exact comment was "The longer the maturity, the more and more gaussian the gamma of your option is and therefore the jump costs you less. Whereas for short term maturities, the gamma converges to a dirac delta function. This implies that if there were a jump, the slippage is much higher for shorter to maturity functions."

Link to the explanation: https://quant.stackexchange.com/a/5976/46192

Winnie
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    What was "$x$?" The Dirac Delta is NOT a function. – Mark Viola Apr 30 '20 at 03:16
  • @MarkViola Their explanation was "The longer the maturity, the more and more gaussian the gamma of your option is and therefore the jump costs you less. Whereas for short term maturities, the gamma converges to a dirac delta function. This implies that if there were a jump, the slippage is much higher for shorter to maturity functions." – Winnie Apr 30 '20 at 03:19
  • $f_n \to \delta$ means that for all $\phi \in C^\infty_c(\Bbb{R})$ (often for $\phi$ in a larger subset, Schwartz space, bounded $C^1$ functions..) $\int_\Bbb{R} \phi(x)f_n(x)dx\to \phi(0)$ as $n\to \infty$ – reuns Apr 30 '20 at 03:20
  • Is it possible to explain that in a more qualitative way? I don't have a pure math background as I am in high finance. – Winnie Apr 30 '20 at 03:22

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The Dirac delta function is not really a function. Roughly speaking, it is a very tall, very narrow function centered on $0$ with area $1$ beneath it. One can often think of it as the "limit" of a series of functions that get taller and taller and narrower and narrower. They could be Gaussians, they could be step functions

$$\delta_n(x)=\begin {cases} n&-\frac 1{2n} \lt x \lt \frac 1{2n}\\ 0&\text{otherwise} \end {cases}$$

so $\delta (x) =0$ whenever $x \neq 0$ but if you integrate from any negative number to any positive number the area is $1$. The usual use is to pick out the value of some other function $f(x)$ at $0$, so $\int_{-a}^a f(x)\delta(x) dx=f(0)$, which you can prove by integrating by parts. I don't know what gamma and slippage are in your use.

Ross Millikan
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  • Thank you so much for explaining it to me. Very kind of you! It was super helpful :) – Winnie Apr 30 '20 at 03:54
  • Gamma is the rate of change or the acceleration amount of delta on an option! Delta is the amount an option price is expected to change given $1 change in the underlying stock. – Winnie Apr 30 '20 at 03:58